數學problem－dvhtlbz的部落格

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數學problem

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3(a)f (x) = kx +12. Find k if f(k) = -7k(b)Solve the equation 5^5-3x= 2^x+21a) solve (2^3x)(3^2y)=9, (5^x)(6^y)=20b) the curve y= a+bx+cx^2 passes through (1,-6), (2,-20), (-2,-12), find the values of a, b and c.2a) find the area of the curve bounded by y=x^2+3, x=0, x=6 and the x axis.b) find the... 顯示更多 3(a)f (x) = kx +12. Find k if f(k) = -7k (b)Solve the equation 5^5-3x= 2^x+2 1a) solve (2^3x)(3^2y)=9, (5^x)(6^y)=20 b) the curve y= a+bx+cx^2 passes through (1,-6), (2,-20), (-2,-12), find the values of a, b and c. 2a) find the area of the curve bounded by y=x^2+3, x=0, x=6 and the x axis. b) find the centroid of the area in (a)

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(3)(a) f(x) = kx + 12 f(k) = k^2 + 12 = -7k k^2 + 7k + 12 = 0 (k + 3)(k + 4) = 0 k = -3 or k = -4 (b) 5^(5-3x) = 2^(x+2) (5-3x)log5 = (x+2)log2 5log5 - 2log2 = (log2 + 3log5)x x = (5log5 - 2log2) / (log2 + 3log5) = 1.206 (1)(a) (2^3x)(3^2y) = 9 => log(2^3x) + log(3^2y) = log9 3xlog2 + 2ylog3 = log9 3log2 x + 2log3 y = log9 y = (log9 – 3log2 x) / 2log3 … (1) (5^x)(6^y) = 20 => log(5^x) + log(6^y) = log20 log5 x + log6 y = log 20 Use (1), log5 x + log6 [(log9 – 3log2 x) / 2log3] = log20 2(log3)(log5) x + (log6)(log9) – 3(log6)(log2) x = 2log(3)log(20) 2(log3)(log5) x – 3(log6)(log2) x = 2(log3)(log20) – (log6)(log9) x = [2(log3)(log20) – (log6)(log9)] / [2(log3)(log5) – 3(log6)(log2)] = -13.96 y = (log9 – 3log2 * -13.96) / [2log(3)] = 14.21 (b) y = a + bx + cx^2 Sub (1,-6), (2,-20), (-2,-12) into the equation respectively -6 = a + b + c … (1) -20 = a + 2b + 4c … (2) -12 = a – 2b + 4c … (3) (2) – (3) => -8 = 4b => b = -2 Sub into (3), -12 = a + 4 + 4c a + 4c = -16 … (4) (1) => a – 2 + c = -6 a + c = -4 … (5) (4) – (5) => 3c = -12 => c = -4 Sub into (5), a – 4 = -4 => a = 0 Therefore y = -2x – 4x^2 (2)(a) Area = ∫ydx from 0 to 6 = ∫(x^2 + 3)dx from 0 to 6 = [(1/3)x^3 + 3x ] from 0 to 6 = (1/3)(216) + 18 = 90 (b) y-ordinate of centroid = ∫ (y/2)ydx / ∫ydx from 0 to 6 (The centroid of each rectangular bar of height y and width dx is at a height of y/2) = (1/2) ∫ y^2 dx / 90 from 0 to 6 = ∫ y^2 dx / 180 from 0 to 6 = ∫ (x^2 + 3)^2 dx / 180 from 0 to 6 = ∫ (x^4 + 6x^2 + 9) dx / 180 from 0 to 6 = [(1/5)x^5 + 2x^3 + 9x] / 180 from 0 to 6 = (0.2*7776 + 2*216 + 54)/180 = 11.34 x-ordinate of centroid = ∫ xydx / ∫ydx from 0 to 6 = ∫ x(x^2 + 3)dx / 90 from 0 to 6 = ∫ (x^3 + 3x)dx / 90 from 0 to 6 = (1/4)x^4 + (3/2)x^2 / 90 from 0 to 6 = (324 + 54)/90 = 4.2 Centroid is (4.2, 11.34)

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